- #1
floyd0117
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I have a problem where I need to figure out the initial velocity vector [itex] \vec{v_0} [/itex] of a projectile, in order for it to land at the final position [itex]\vec{r_f} = x_f\hat{x} + y_f\hat{y} + z_f\hat{z}[/itex], from initial position [itex]\vec{r_0}[/itex].
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The only knowns in the problem are [itex]\vec{r_0}[/itex] and [itex]\vec{r_f}[/itex]. Air resistance is neglected, so the the components of the net force on the projectile are
[itex]m\ddot{x} = 0[/itex]
[itex]m\ddot{y} = 0[/itex]
[itex]m\ddot{z} = -mg[/itex]
So really we can choose any launch angle [itex]\phi[/itex], and find the necessary [itex]|\vec{v}|[/itex], or the other way around, to land us at [itex]\vec{r_f}[/itex]. I think it sounds easier to choose a [itex]\phi[/itex] and then find [itex]|\vec{v}|[/itex]. So, I examine the limiting cases...
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Let's say [itex]d[/itex] is the the distance between the initial and final positions in the [itex]x[/itex]-[itex]y[/itex] plane, that is;
[itex]d = |x_f\hat{x} + y_f\hat{y}|[/itex]
and that [itex]h[/itex] is the desired final height, [itex]h = z_f[/itex].
Then the angle [itex]\theta[/itex] measured form the [itex]x[/itex]-[itex]y[/itex] plane to a line connecting [itex](x_0, y_0 ,z_0)[/itex] to [itex](x_f, y_f, z_f)[/itex] is smiply
[itex]\theta = \arctan{\dfrac{h}{d}}[/itex]
So, our limiting cases are:
[itex]\phi \rightarrow \theta; |\vec{v_0}| \rightarrow \infty[/itex]
[itex]\phi \rightarrow \dfrac{\pi}{2}; |\vec{v_0}| \rightarrow \infty[/itex]
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So I can choose any angle between [itex]\dfrac{\pi}{2}[/itex] and [itex]\theta[/itex], though angles close to those values will necessitate a very large initial velocity. My question is, how do I go from here, to determining [itex]|\vec{v_0}|[/itex]? If I choose a [itex]\phi[/itex], how do I find a velocity that will get me to [itex]\vec{r_f}[/itex]? It would seem that I need some function of [itex]v_0[/itex] in terms of both [itex]\phi[/itex] (known, after choosing), and [itex]\vec{r_f}[/itex]. Am I severely over thinknig this?
___
The only knowns in the problem are [itex]\vec{r_0}[/itex] and [itex]\vec{r_f}[/itex]. Air resistance is neglected, so the the components of the net force on the projectile are
[itex]m\ddot{x} = 0[/itex]
[itex]m\ddot{y} = 0[/itex]
[itex]m\ddot{z} = -mg[/itex]
So really we can choose any launch angle [itex]\phi[/itex], and find the necessary [itex]|\vec{v}|[/itex], or the other way around, to land us at [itex]\vec{r_f}[/itex]. I think it sounds easier to choose a [itex]\phi[/itex] and then find [itex]|\vec{v}|[/itex]. So, I examine the limiting cases...
____
Let's say [itex]d[/itex] is the the distance between the initial and final positions in the [itex]x[/itex]-[itex]y[/itex] plane, that is;
[itex]d = |x_f\hat{x} + y_f\hat{y}|[/itex]
and that [itex]h[/itex] is the desired final height, [itex]h = z_f[/itex].
Then the angle [itex]\theta[/itex] measured form the [itex]x[/itex]-[itex]y[/itex] plane to a line connecting [itex](x_0, y_0 ,z_0)[/itex] to [itex](x_f, y_f, z_f)[/itex] is smiply
[itex]\theta = \arctan{\dfrac{h}{d}}[/itex]
So, our limiting cases are:
[itex]\phi \rightarrow \theta; |\vec{v_0}| \rightarrow \infty[/itex]
[itex]\phi \rightarrow \dfrac{\pi}{2}; |\vec{v_0}| \rightarrow \infty[/itex]
___
So I can choose any angle between [itex]\dfrac{\pi}{2}[/itex] and [itex]\theta[/itex], though angles close to those values will necessitate a very large initial velocity. My question is, how do I go from here, to determining [itex]|\vec{v_0}|[/itex]? If I choose a [itex]\phi[/itex], how do I find a velocity that will get me to [itex]\vec{r_f}[/itex]? It would seem that I need some function of [itex]v_0[/itex] in terms of both [itex]\phi[/itex] (known, after choosing), and [itex]\vec{r_f}[/itex]. Am I severely over thinknig this?
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